Newbie question

[Del Rio]

Ok I know someone on this site must be a math genious. So here goes. I am in charge of making a work list that will run for six months.

There are 12 guys, who will be working in groups of 2. And they will need to be in a different group every month (not grouped with the same people twice.)

That is fine and dandy, but the hospital has 6 floors, so not only do I need every group to be different each month I need each person to get the opportunity to work on each floor.

------------------------------------------------------------

I was thinking I would let each letter of the alphabet equal a person so you have

A B C D E F G H I J K L------------REPRESENTS EACH PERSON 12 TOTAL.

So if you draw a grid out 6months down the side- the floors over the top. and you start grouping the letters together.....AB, CD......so on it equals a huge pain. I have gotten 5 months done and 5 floors where no one is grouped together twice, and no one is on the same floor. But it is looking like there is only one possible solution to the problem.

 

[Denis]

Hmmm...looking at it with 8 guys, 4 months, 4 floors:

Write 'em down like this:
ab 
ac bc 
ad bd cd 
ae be ce de 
af bf cf df ef 
ag bg cg dg eg fg
ah bh ch dh eh fh gh

1: enter 1st 4 a-combos diagonally, and
2: enter last 4 h-combos diagonally:

ab ?? ?? gh
?? ac fh ??
?? eh ad ??
dh ?? ?? ae

So now we can't enter any more a-combos or h-combos, 
so remove a-column, h-row:

bc 
bd cd 
be ce de 
bf cf df ef 
bg cg dg eg fg

...continue similarly, eliminating the b-column and g-row

That's just an idea; may work; too lazy to finish off!
And VERY happy to finally being able to use the "code" button!!

 

[Denis]

Well, the "genius" you're looking for is sure not me;
just realised my "good looking" strategy will NOT work!

BUT I was assuming a person could work on a certain floor only once:
that's not so, right?

Now I'm confused:
can you show me a "4 persons, 2 months, 2 floors" schedule?

 

[Gene]

I agree with Denis' assumption. They can't work the same floor twice or there will be a floor they don't work on.
He doesn't say why his method fails. He had me convinced. Just 'cause it doesn't work for 8-4-4 doesn't prove it won't work for 12-6-6. I started with the simplest 4-2-2 but the first week has to be AB,CD (or they can be swapped so they are) and you get

AB CD    or   AB CD
C? --         ?D --

and there is nothing to put in the ? that works so it is impossible. If his method doesn't work for 12-6-6 (that may be the smallest that can be done) I think it proves impossibility for that too. I like his logic.

BTW, What's wrong with there being only one solution?
-----------------
Gene

PS Edit. Okay, I'm convinced. It doesn't work as written. Back to the scratch-pad.

 

[Denis]

A 8-4-4 solution:
ab-dg-ef-ch
eg-ac-bh-df
cf-eh-ad-bg
dh-bf-cg-ae

Sorta works using my original plan.

 

[Gene]

but

2: enter last 4 h-combos diagonally...

continue similarly, eliminating the b-column and g-row

But
ab-dg-ef-ch
eg-ac-bh-df
cf-eh-ad-bg
dh-bf-cg-ae

violates both those rules "as written". The original plan looked so good and clean and logical that I regretted not thinking of it till you 'fessed up.

BTW Have we lost DEL RIO :?:
Is this now another pillow problem to be thought about only on long winter nigts when I can't sleep :?:
------------------
Gene

 

[Denis]

As before, write 'em down like this: 
ab 
ac bc 
ad bd cd 
ae be ce de 
af bf cf df ef 
ag bg cg dg eg fg 
ah bh ch dh eh fh gh 

Remove some this way (leaves 16):
ab 
ac xx
ad xx xx 
ae xx xx xx 
xx bf cf df ef 
xx bg cg dg eg xx xx 
xx bh ch dh eh xx xx xx 

Enter in grid this way (numbers are row numbers):
1ab 
2ac xx
3ad xx xx 
4ae xx xx xx 
xx 4bf 3cf 2df 1ef 
xx 3bg 4cg 1dg 2eg xx xx 
xx 2bh 1ch 4dh 3eh xx xx xx 

So there is some pattern!

ab-dG-eF-ch 
eG-ac-bh-dF
cF-eh-ad-bG
dh-bF-cG-ae 

The a's and h's = 2 diagonals
The f's also diagonal, in pairs
The g's also diagonal, in pairs

I think a 12-6-6 can be made; start with a 8-4-4, then build around it;
I'm pretty sure the "diagonal arrangements" will remain same...extended.

 

[Del Rio]

I apologize for the lack of response. I have been really busy.

I appreciate the help. The first suggestion will not work, as you realized. I discovered that as well. I used Pascal's Triangle as a map trying to find a formula.

I ended up getting frustrated and after three days of plugging in letters I found a correct solution.

Four columns and four rows is the easy part it is column five and row five that begins to get hard. Row 6 and column 6 are darn near impossible. That is why I was wondering if there was a formula or a solution like the one Denis suggested.

When I was mapping out my selections on the triangle of options I was noticing a distinct pattern. I noticed that I wasn't pulling any from a certain section of the choices, I could not figure out how to implement a design that would pull those into the solution.

So I started plugging and moving and using trial and error, mostly out of frustration. Eventually I came up with this as a solution....

AB CD EF GH IJ KL
EK AG CI BJ FL DH
CF BE AD IL HK GJ
DL FJ BH AK CG EI
GI HL JK DF AE BC
HJ IK GL CE BD AF

Keeping in mind that it was just logic and guessing for about two solid hours. I KNOW there has to be a way to simplify the process and I am still messing with it trying to figure it out. This puts the task to rest for me, but has opened a new mission.

Any other ideas?

*note* Yeah a person can not work on the same floor twice, he also cannot be grouped in the same group more then once, and he is only allowed to work once per month.

 

[Del Rio]

There is nothing wrong with there being only one solution, but given the ammount of variation I was just assuming there would be more.

I guess the main problem with there being only one solution is with the lack of ability to stream line the process I would be basically guessing what letters to place where. If there was only one answer it would probably take me forever to find it.

Besides I was running out of paper

 

[Gene]

Glad you're still here. Even more glad to hear you have a solution. And gladest of all to know that my (useless) thought was not for naught. I'm still looking for a "magic" Denis type recipe.
Its similar to the old puzzle of laying out a square with the honor cards so no two values or suits are in the same row or column but I don't recall if there was a method for that or if it was just trial and error.
I'm not sure if I agree "there is only one solution" 'cause even if you freeze the first row there are 19 more solutions with permutations of the other 5 rows. I know reflections and rotations don't count but...
They all use the same 36 pairs so I guess they would be considered the same but they do LOOK different.
Thanks for a different challenge than "how do I substitute for y" type questions.
--------------------
Gene